Let \(\max \lbrace  1 , p_{+} \rbrace  < p_0 < \frac{n}{\alpha }\) . Given \(f \in H^{p(.)} \cap L^{p_0}(\mathbb {R}^{n}),\)  from Theorem 4.6 in [1]}
we have that there exist a sequence of nonnegative numbers \(\lbrace \lambda _j\rbrace _{j=1}^{\infty }\) , a sequence of cubes \(Q_j = Q(z_j,r_j)\) 
centered at \(z_j\)  with side length \(r_j\)  and \((p(.), p_0, d)\)  atoms \(a_j\)  supported on \(Q_j\) , satisfying
\(\mathcal {A}\left( \lbrace \lambda _j\rbrace _{j=1}^{\infty }, \lbrace Q_j\rbrace _{j=1}^{\infty }, p(.) \right) \le c \Vert f \Vert _{H^{p(.)}},\) 
such that \(f\)  can be decomposed as \(f = \sum _{j \in \mathbb {N}} \lambda _j a_j\) , where the convergence is in \(H^{p(.)}\)  and in \(L^{p_0}\)  (for the converge in \(L^{p_0}\)  see Theorem 5 in [2]}, the same argument works for \(f \in H^{p(.)} \cap L^{p_0}(\mathbb {R}^{n})\) ).
We will study the behavior of \(T_{\alpha }\)  on atoms. Define \(D=\max \limits _{1\le i\le m,\left\Vert x\right\Vert \le 1}\left\lbrace  \left\Vert A_{i}(x)\right\Vert \right\rbrace  .\) 
Fix \(j \in \mathbb {N}\) , let \(a_j\)  be an \((p(.), p_0, d)\) -atom supported on a cube \(Q_j = Q(z_j, r_j)\) , for each \(1 \le i \le m\)  let \(Q_{ji}^{\ast } = Q(A_i z_j, 4Dr_j)\) . Proposition 1 in [3]} gives that \(T_{\alpha }\)  is bounded from \(L^{p_{0}}\left( \mathbb {R}^{n}\right) \)  into
\(L^{q_{0}}\left( \mathbb {R}^{n}\right)\)  for \(\frac{1}{q_{0}}=\frac{1}{p_{0}}-\frac{\alpha }{n},\)  thus
\(\left\Vert T_{\alpha } a_j \right\Vert _{L^{q_{0}}(Q_{j i}^{\ast })}\lesssim \left\Vert a\right\Vert _{p_{0}}\lesssim \frac{\left|Q_j \right|^{\frac{1}{p_{0}}}}{\left\Vert \chi _{Q_j }\right\Vert _{p(.)}}\lesssim \frac{\left|Q_{ji}^{\ast } \right|^{\frac{1}{q_{0}}}}{\left\Vert \chi _{Q_{j i}^{\ast } }\right\Vert _{q(.)}},\) 
where the last inequality follows from lemma 4. So if \(\left\Vert T_{\alpha }a_j \right\Vert _{L^{q_{0}}(Q_{ji}^{\ast })}\ne 0\)  we get
\(1\lesssim \frac{\left|Q_{ji}^{\ast } \right|^{\frac{1}{q_{0}}}}{\left\Vert T_{\alpha } a_j \right\Vert _{L^{q_{0}}(Q_{ji}^{\ast })}\left\Vert \chi _{Q_{j i}^{\ast }}\right\Vert _{q(.)}}. \) 
We denote \(k(x,y)=\left|x-A_{1}y\right|^{-\alpha _{1}}...\left|x-A_{m}y\right|^{-\alpha _{m}}.\)  In view of the moment condition of \(a_j\)  we have
\(T_{\alpha } a_j(x)=\int \limits _{Q_j}k(x,y)a_j(y)dy=\int \limits _{Q_j}\left( k(x,y)-q_{d, j}\left(x,y\right) \right) a_j(y)dy,\) 
where \(q_{d, j}\)  is the degree \(d\)  Taylor polynomial of the function \(y\rightarrow k(x,y)\)  expanded around \(z_j\) . By the standard estimate of the
remainder term of the taylor expansion, there exists \(\xi \)  between \(y\)  and
\(z_j\)  such that
\(\left|k(x,y)-q_{d, j}\left( x,y\right) \right|\lesssim \left|y-z_j \right|^{d+1}\sum \limits _{k_{1}+...+k_{n}=d+1}\left|\frac{\partial ^{d+1}}{\partial y_{1}^{k_{1}}...\partial y_{n}^{k_{n}}}k(x,\xi )\right|\) 
\(\lesssim \left|y-z_j \right|^{d+1}\left(\prod \limits _{i=1}^{m}\left|x-A_{i}\xi \right|^{-\alpha _{i}}\right) \left( \sum \limits _{l=1}^{m}\left|x-A_{l}\xi \right|^{-1}\right) ^{d+1}.\) 
Now we decompose \(\mathbb {R}^{n} = \bigcup _{i=1}^{m} Q_{ji}^{\ast } \cup R_j\) , where \(R_j = \left( \bigcup _{i=1}^{m} Q_{ji}^{\ast } \right)^{c}\) , at the same time we decompose \(R_j = \bigcup _{k=1}^{m} R_{jk}\)  with
\(R_{jk} = \lbrace  x \in R_{j} : |x - A_k z_j| \le |x - A_i z_j| \,\, for \,\, all \,\, i \ne k \rbrace .\) 
If \(x \in R_{j}\)  then \(|x - A_i z_j| \ge 2Dr_j\) , since \(\xi \in Q_j\)  it follows that \(|A_i z_j - A_i \xi | \le D r_j \le \frac{1}{2} |x - A_i z_j|\)  so
\(|x - A_i \xi | = |x - A_i z_j + A_i z_j - A_i \xi | \ge |x - A_i z_j| - |A_i z_j - A_i \xi | \ge \frac{1}{2} |x - A_i z_j|.\) 
If \(x \in R_j\) , then \(x \in R_{jk}\)  for some \(k\)  and since \(\alpha _{1}+...+\alpha _{m} = n - \alpha \)  we obtain
\(\left|k(x,y)-q_{d, j}\left( x,y\right) \right|\lesssim \left|y-z_j \right|^{d+1}\left( \prod \limits _{i=1}^{m}\left|x-A_{i}z_j \right|^{-\alpha _{i}}\right) \left(\sum \limits _{l=1}^{m}\left|x-A_{l}z_j \right|^{-1}\right) ^{d+1}\) 
\(\lesssim r_{j}^{d+1}\left|x-A_{k}z_j \right|^{-n+\alpha -d-1},\) 
this inequality allow us to conclude that
\(\left|T_{\alpha }a_j(x)\right|&\lesssim &\left\Vert a_j \right\Vert _{1}r_{j}^{d+1}\left|x-A_{k}z_j\right|^{-n+\alpha -d-1} \\&\lesssim &\left|Q_j \right|^{1-\frac{1}{p_{0}}}\left\Vert a_j \right\Vert _{p_{0}}r_{j}^{d+1}\left|x-A_{k}z_j\right|^{-n+\alpha -d-1}\\&\lesssim &\frac{r_{j}^{n+d+1}}{\left\Vert \chi _{Q_j}\right\Vert _{p(.)}}\left|x-A_{k}z_j \right|^{-n+\alpha -d-1} \\&\lesssim &\frac{\left( M_{\frac{\alpha n}{n+d+1}}\left( \chi _{Q_j}\right)(A_{k}^{-1}x)\right) ^{\frac{n+d+1}{n}}}{\left\Vert \chi _{Q_j}\right\Vert _{p(.)}}, \,\,\,\,\, if \,\, x \in R_{jk}.\) 
Since \(f = \sum _{j=1}^{\infty } \lambda _j a_j\)  in \(L^{p_0}\)  and \(T_{\alpha }\)  is an \(L^{p_0}-L^{\frac{n p_0}{n- \alpha p_0}}\)  bounded operator, we have that \(|T_{\alpha } f (x)| \le \sum _{j=1}^{\infty } \lambda _j |T_{\alpha } a_j(x)|\) . So
\(|T_{\alpha } f (x)| \le \sum _{j=1}^{\infty } \lambda _j |T_{\alpha } a_j(x)|= \sum _{j=1}^{\infty } \left(\chi _{\bigcup _{i=1}^{m} Q_{ji}^{\ast }}(x) + \chi _{R_j}(x) \right) \lambda _j |T_{\alpha } a_j(x)|\) 
\(\lesssim \sum _{j=1}^{\infty } \sum _{i=1}^{m} \chi _{Q_{ji}^{\ast }}(x) \lambda _j |T_{\alpha } a_j(x)| + \sum _{j=1}^{\infty } \sum _{k=1}^{m} \chi _{R_{jk}}(x) \lambda _j |T_{\alpha } a_j(x)|\) 
\(\lesssim \sum _{j=1}^{\infty } \sum _{i=1}^{m} \chi _{Q_{ji}^{\ast }}(x) \lambda _j |T_{\alpha } a_j(x)|\) 
\(+ \sum _{j=1}^{\infty } \sum _{k=1}^{m} \chi _{R_{jk}}(x) \lambda _j \frac{\left( M_{\frac{\alpha n}{n+d+1}}\left( \chi _{Q_j}\right)(A_{k}^{-1}x)\right) ^{\frac{n+d+1}{n}}}{\left\Vert \chi _{Q_j}\right\Vert _{p(.)}}= I + II\) 
To study \(I,\)  if \(\left\Vert T_{\alpha }a_{j}\right\Vert _{L^{q_{0}}(Q_{ji}^{\ast })}\ne 0,\)  we apply (REF ) to
obtain, since \(\underline{q}\le 1,\)
\(\left\Vert I\right\Vert _{q(.)}\lesssim \sum \limits _{i=1}^{m}\left\Vert \sum \limits _{j=1}^{\infty }\lambda _{j}\chi _{Q_{ji}^{\ast }}\left|T_{\alpha }a_{j} \right|\right\Vert _{q(.)}\lesssim \sum \limits _{i=1}^{m}\left\Vert \sum \limits _{j=1}^{\infty }\frac{\lambda _{j}\chi _{Q_{ji}^{\ast }}\left|T_{\alpha } a_{j} \right|\left|Q_{ji}^{\ast }\right|^{\frac{1}{q_{0}}}}{\left\Vert T_{\alpha }a_{j}\right\Vert _{L^{q_{0}}(Q_{ji}^{\ast })}\left\Vert \chi _{Q_{ji}^{\ast }}\right\Vert _{q(.)}}\right\Vert _{q(.)}\) 
\(\lesssim \sum \limits _{i=1}^{m}\left\Vert \left\lbrace  \sum \limits _{j=1}^{\infty }\left(\frac{\lambda _{j}\chi _{Q_{ji}^{\ast }}\left|T_{\alpha } a_{j}\right|\left|Q_{ji}^{\ast }\right|^{\frac{1}{q_{0}}}}{\left\Vert T_{\alpha }a_{j} \right\Vert _{L^{q_{0}}(Q_{ji}^{\ast })}\left\Vert \chi _{Q_{ji}^{\ast }}\right\Vert _{q(.)}}\right) ^{\underline{q}}\right\rbrace  ^{\frac{1}{\underline{q}}}\right\Vert _{q(.)},\) 
now we take \(p_{0}\)  near \(\frac{n}{\alpha }\)  such that \(\delta =\frac{1}{q_{0}}\)  satisfies
the hypothesis of Lemma 4.11 in [1]} to get
\(\lesssim \sum _{i=1}^{m} \mathcal {A}\left( \left\lbrace  \lambda _{j}\right\rbrace  _{j=1}^{\infty },\left\lbrace Q_{j i}^{\ast }\right\rbrace  _{j=1}^{\infty },q(.)\right)\) 
Lemma 5 gives
\(\lesssim m \mathcal {A}\left( \left\lbrace  \lambda _{j}\right\rbrace  _{j=1}^{\infty },\left\lbrace Q_{j}\right\rbrace  _{j=1}^{\infty },q(.)\right)\) 
now from Lemma 7,
\(\lesssim \mathcal {A}\left( \left\lbrace  \lambda _{j}\right\rbrace  _{j=1}^{\infty },\left\lbrace Q_{j}\right\rbrace  _{j=1}^{\infty },p(.)\right) \lesssim \left\Vert f\right\Vert _{H^{p(.)}}.\) 
To study \(II,\)  we observe that
\(\left\Vert II\right\Vert _{q(.)} = \left\Vert \sum \limits _{j=1}^{\infty }\sum \limits _{k=1}^{m}\lambda _{j}\chi _{R_{jk}}\left( .\right)\frac{\left( M_{\frac{\alpha n}{n+d+1}}\left( \chi _{Q_{j}}\right)(A_{k}^{-1} .)\right) ^{\frac{n+d+1}{n}}}{\left\Vert \chi _{Q_j}\right\Vert _{p(.)}}\right\Vert _{q(.)} \\\lesssim m \left\Vert \left\lbrace  \sum \limits _{j=1}^{\infty }\lambda _{j}\frac{\left( M_{\frac{\alpha n}{n+d+1}}\left( \chi _{Q_{j}}\right)(A_{k}^{-1} .)\right) ^{\frac{n+d+1}{n}}}{\left\Vert \chi _{Q_{j}}\right\Vert _{p(.)}}\right\rbrace  ^{\frac{n}{n+d+1}}\right\Vert _{\frac{n+d+1}{n}q(A_{k}.)}^{\frac{n+d+1}{n}}\) 
\(\lesssim \left\Vert \left\lbrace  \sum \limits _{j=1}^{\infty }\lambda _{j}\frac{\chi _{Q_{j}}}{\left\Vert \chi _{Q_{j}}\right\Vert _{p(.)}}\right\rbrace  ^{\frac{n}{n+d+1}}\right\Vert _{\frac{n+d+1}{n}p(A_{k}.)}^{\frac{n+d+1}{n}}=\left\Vert \sum \limits _{j=1}^{\infty }\lambda _{j}\frac{\chi _{Q_{j}}}{\left\Vert \chi _{Q_{j}}\right\Vert _{p(.)}}\right\Vert _{p(.)}\) 
\(\lesssim \mathcal {A}\left( \left\lbrace  \lambda _{j}\right\rbrace  _{j=1}^{\infty },\left\lbrace Q_{j}\right\rbrace  _{j=1}^{\infty },p(.)\right) \lesssim \left\Vert f\right\Vert _{H^{p(.)}},\) 
where the second inequality follows from Lemma 6, since \(\frac{n+d+1}{n}\underline{q}>1,\)  the third inequality follows from Remark 4.4 in [1]}
and the second equality follows since \(p(A_{k}x)\equiv p(x).\)  Thus
\(\Vert  T_{\alpha } f \Vert _{q(.)} \lesssim \Vert  f\Vert _{H^{p(.)}}\) 
for all \(f \in H^{p(.)} \cap L^{p_0}(\mathbb {R}^{n})\) , so the theorem follows from the density of \(H^{p(.)} \cap L^{p_0}(\mathbb {R}^{n})\)  in \(H^{p(.)}(\mathbb {R}^{n})\) .
